Given:
A circle C with radius R and center (x,y),
A point P at (q,r) some distance d away from the circle at its center line, and at a certain height above that center line h, and
An angle of displacement a,
Find the distance from the point P to the circle for both cases: when the angle $a$ is zero and when the angle $a$ is greater than zero.
Looking forward to whatever solution or even advice someone might be able to give me in order to solve this! Thanks in advance!
Best Answer
So, we need to find out the distance $PA$ as a function of $\alpha$, including the special case $\alpha =0$. It is a quite interesting and challenging problem.
Let $\beta=∠ACB$ and apply the tangent formula to the right triangle APD,
$$\tan\alpha = \frac{AD}{PD}=\frac{R\sin\beta - h}{R+d-R\cos\beta}$$
Rearrange above equation to express $\beta$ in terms of $\alpha$,
$$R\sin(\beta+\alpha)=(R+d)\sin\alpha + h\cos\alpha$$
$$\beta = \sin^{-1} u -\alpha \tag{1}$$
where,
$$u = \left(1+\frac{d}{R}\right)\sin\alpha + \frac{h}{R}\cos\alpha \tag{2}$$
Next, apply the cosine formula to the right triangle APD to write the distance $PA$ as,
$$PA(\alpha) =\frac{R+d-R\cos\beta}{\cos\alpha}\tag{3} $$
With (1), we evaluate $R\cos\beta$ in above expression to get,
$$R\cos\beta = R\sqrt{1-u^2} \cos\alpha + (R+d)\sin^2\alpha + h\cos\alpha\sin\alpha \tag{4}$$
Plug (2) into (4) and then into into (3), we arrive at the following result,
$$PA(\alpha) =(R+d)\cos\alpha -h\sin\alpha - \sqrt{R^2-[ (R+d)\sin\alpha + h\cos\alpha ]^2} \tag{5}$$
As seen, the distance $PA$ varies with the angle $\alpha$ in a non-trivial way.
Now, consider the special case of $\alpha = 0$. The general result (5) then simplifies greatly,
$$PA(0) = R + d -\sqrt{R^2-h^2}$$